Vectors and Projectiles
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Part D: ProblemSolving
56. In the Vector Addition Lab, Anna starts at the classroom door and walks:

2.0 meters, West

12.0 meters, North,

31.0 meters, West,

8.0 meters, South

3.0 meters, East
Using either a scaled diagram or a calculator, determine the magnitude and direction of Anna's resulting displacement.
Answer: 30.3 meters, 172 degrees
To insure the most accurate solution, this problem is best solved using a calculator and trigonometric principles. The first step is to determine the sum of all the horizontal (eastwest) displacements and the sum of all the vertical (northsouth) displacements.
Horizontal: 2.0 meters, West + 31.0 meters, West + 3.0 meters, East = 30.0 meters, West
Vertical: 12.0 meters, North + 8.0 meters, South = 4.0 meters, North
The series of five displacements is equivalent to two displacements of 30 meters, West and 4 meters, North. The resultant of these two displacements can be found using the Pythagorean theorem (for the magnitude) and the tangent function (for the direction). A nonscaled sketch is useful for visualizing the situation.
Applying the Pythagorean theorem leads to the magnitude of the resultant (R).
R^{2} = (30.0 m)^{2} + (4.0 m)^{2} = 916 m^{2}
R = Sqrt(916 m^{2})
R = 30.3 meters
The angle theta in the diagram above can be found using the tangent function.
tangent(theta) = opposite/adjacent = (4.0 m) / (30.0 m)
tangent(theta) = 0.1333
theta = invtan(0.1333)
theta = 7.59 degrees
This angle theta is the angle between west and the resultant. Directions of vectors are expressed as the counterclockwise angle of rotation relative to east. So the direction is 7.59 degrees short of 180 degrees. That is, the direction is ~172 degrees.
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57. In a grocery store, a shopper walks 36.7 feet down an aisle. She then turns left and walks 17.0 feet straight ahead. Finally, she turns right and walks 8.2 feet to a final destination. (a) Determine the magnitude of the overall displacement. (b) Determine the direction of the displacement vector relative to the original line of motion.
Answer: (a) 48.0 feet; (b) 21 degrees from the original line of motion
This problem is best approached using a diagram of the physical situation. The three displacements are shown in the diagram below on the left. Since the three displacements could be done in any order without effecting the resulting displacement, these three legs of the trip are conveniently rearranged in the diagram below on the right.
Now it is obvious from the diagram on the right that the three displacement vectors are equivalent to two perpendicular displacement vectors of 44.9 feet and 17 feet. These two vectors can be added together and the resultant can be drawn from the starting location to the final location. A sketch is shown below.
Since these displacement vectors are at right angles to each other, the magnitude of the resultant can be determined using the Pythagorean theorem. The work is shown below.
R^{2} = (44.9 ft)^{2} + (17.0 ft)^{2} = 2305 ft^{2}
R = Sqrt(2305 ft^{2})
R = 48.0 feet
The angle theta in the diagram above can be found using the tangent function.
tangent(theta) = opposite/adjacent = (17.0 ft) / (44.9 ft)
tangent(theta) = 0.3786
theta = invtan(0.3786)
theta = 20.7 degrees
This is the angle which the resultant makes with the original line of motion (the 36.7 ft displacement vector).
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58. A hiker hikes 12.4 km, south. The hiker then makes a turn towards the southeast and finishes at the final destination. The overall displacement of the twolegged trip is 19.7 km at 309 degrees . Determine the magnitude and direction of the second leg of the trip.
Answer: 12.7 km, 347 degrees
Like the previous problem (and most other problems in physics), this problem is best approached using a diagram. The first displacement is due South and the resulting displacement (at 309 degrees) is somewhere in the fourth quadrant. (It is in the fourth quadrant because 309 degrees lies between 270 degrees or due South and 360 degrees or due East.) For communication sake, we will refer to the first displacement as A and the second displacement as B. Note that A + B = R. Since the magnitude and direction of the resultant is known, the x and ycomponents can be determined using trigonometric functions. Since the angle of 309 degrees is expressed as a counterclockwise angle of rotation with due East, it can be used as the Theta in the equation.
R_{x} = R•cos(theta) = 19.7 km • cos(309 deg) = 12.398 km
R_{y} = R•sin(theta) = 19.7 km • sin(309 deg) = 15.310 km (the "" means South)
Whatever the magnitude and direction of B is, it must add on to vector A in order to give a southward displacement of 15.310 km and a eastward displacement of 12.398 km. This could be expressed by mathematical equations as
A_{x} + B_{x} = 12.398 km, east
A_{y} + B_{y} = 15.310 km, south
But A_{x} is 0 km and A_{y} is 12 km, South. By substituting these two values into the above equations, the values for the x and ycomponents of the unknown displacement can be determined:
B_{x} = 12.398 km, east and B_{y} = 2.910 km, south
Knowing the B_{x} and B_{y} components will allow us to determine the magnitude and direction of B. Another diagram would help in visualizing the situation. The magnitude of B can be found using the Pythagorean theorem. B is the hypotenuse of a right triangle with legs of 12.398 km and 2.910 km. The sum of the squares of the sides is equal to the square of the hypotenuse.
B^{2} = (12.398 km)^{2} = (2.910 km)^{2}
B^{2} = 162.179 km^{2}
B = Sqrt(162.179 km^{2})
B = 12.735 km
The direction of B is close to 360 degrees. As shown in the diagram, it is less than 360 degrees by an amount equal to theta. The angle theta can be determined using the tangent function and the length of the two sides of the right triangle.
Tangent (Theta) = (2.910 km) / (12.398 km)
Tangent (Theta) = 0.2347
Theta = Invtan (0.2347)
Theta = 13.2 degrees
The direction of B is 360 degrees  13.2 degrees = ~347 degrees.
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59. A boat heads straight across a river which is 100. meters wide. For the following two combinations of boat velocities and current velocities, determine the resultant velocity, the time required to cross the river, and the distance traveled downstream.
a.
Given:
Boat velocity = 10.0 m/s, East
River velocity = 4.0 m/s, North
Calculate:
Resultant Vel. (mag. & dir'n): 11 m/s, 22 deg.
Time to cross river: 10.0 s
Distance traveled downstream: 40. m

b.
Given:
Boat velocity = 8.0 m/s, East
River velocity = 5.0 m/s, South
Calculate:
Resultant Vel. (mag. & dir'n): 9.4 m/s, 328 deg.
Time to cross river: 13 s
Distance traveled downstream: 63 m

Answer: See table above.
The two velocity vectors (boat and river) are directed perpendicular to each other. They can be added using the Pythagorean theorem. The direction is found using the tangent function; it is expressed as the counterclockwise angle of rotation from due East. The time to cross the river is dependent upon the width of the river and the boat velocity. And the distance downstream is dependent upon the time that the boat is moving and the speed at which it moves downstream  the river velocity.
a.
Resultant Vel. (mag. & dir'n):
R^{2} = (10.0 m/s)^{2} + (4.0 m/s)^{2}
R = SQRT ( (10.0 m/s)^{2} + (4.0 m/s)^{2} )
R = SQRT(116 m^{2}/s^{2}) = 11 m/s (rounded from 10.8 m/s)
dir'n = invtan(4.0/10.0) = 22 degrees
Time to cross river:
d = v*t > t = d/v
t = (100. m)/(10.0 m/s) = 10.0 s
Distance traveled downstream:
d = v*t = (4.0 m/s)*(10.0 s) = 40. m

b.
Resultant Vel. (mag. & dir'n):
R^{2} = (8.0 m/s)^{2} + (5.0 m/s)^{2}
R = SQRT ( (8.0 m/s)^{2} + (5.0 m/s)^{2} )
R = SQRT(89 m^{2}/s^{2}) = 9.4 m/s (rounded from 9.43 m/s)
dir'n = 360 deg.  invtan(5.0/8.0) = 328 degrees
Time to cross river:
d = v*t > t = d/v
t = (100. m)/(8.0 m/s) = 13 s (rounded from 12.5 s)
Distance traveled downstream:
d = v*t = (5.0 m/s)*(12.5 s) = ~63 m

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60. The di
agram at the right depicts a horizontallylaunched projectile leaving a cliff of height y with a horizontal velocity (v_{ix}) and landing a distance x from the base of the cliff. Express your understanding of projectile kinematics by filling in the blanks in the table below. To simplify the calculations, use the approximated value for the acceleration of gravity of 10 m/s/s.

v_{ix}
(m/s)

y
(m)

t
(s)

x
(m)

a.

15.0 m/s

20.0 m

2.00

30.0

b.

15.0 m/s

45.0

3.00 s

45.0

c.

15.0

45.0 m

3.00

45.0 m

d.

12.0

31.3

2.50 s

30.0 m

e.

17.2

74.0 m

3.85

66.0 m


Answer: See table above.
The solutions to all five of these projectile problems involve the use of kinematic equations and an appropriate problemsolving strategy. The kinematic equations and their use in projectile problems are listed and discussed elsewhere. The basic idea of the strategy is to identify three kinematic variables for either the horizontal motion or for the vertical motion. Once , three quantities in one direction is known, all other quantities in that direction can be found (or the time of flight can be found). Often, the time is then used with kinematic quantities for the second dimension in order to determine all other unknown quantities for that dimension.
In each of these problems, it is known that a_{x} = 0 m/s/s, a_{y} = 10 m/s/s, and v_{iy} = 0 m/s. When these three knowns are combined with the other given knowns the following answers are obtained:

Answers

Method

a.

t = 2.0 s and x = 30.0 m

Use y, v_{iy}, and a_{y} to calculate t; then use t, v_{ix} and a_{x} to calculate x.

b.

y = 45.0 m and x = 45.0 m

Use t, v_{ix} and a_{x} to calculate x; and use t, v_{iy}, and a_{y} to calculate y.

c.

v_{ix} = 15.0 m/s and t = 3.00 s

Use y, v_{iy}, and a_{y} to calculate t; then use t, x and a_{x} to calculate v_{ix}.

d.

v_{ix} = 12.0 m/s and y = 31.3 m

Use t, x and a_{x} to calculate v_{ix}; and use t, v_{iy}, and a_{y} to calculate y.

e.

v_{ix} = 17.2 m/s and t = 3.85 s

Use y, v_{iy}, and a_{y} to calculate t; then use t, x and a_{x} to calculate v_{ix}.

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61. The launch velocity and angle is given for three different projectiles. Use trigonometric functions to resolve the velocity vectors into horizontal and vertical velocity components. Then use kinematic equations to determine the time that the projectile is in the air, the height to which it travels (when it is at its peak), and the horizontal distance that it travels. (To simplify the calculations, use an acceleration of gravity value of 10 m/s/s.)
a.
Given:
Launch Vel. = 30.0 m/s
Launch angle = 30.0 degrees

b.
Given:
Launch Vel. = 30.0 m/s
Launch angle = 45.0 degrees

c.
Given:
Launch Vel. = 30.0 m/s
Launch angle = 50.0 degrees

Calculate:
v_{ix} = 26.0 m/s
v_{iy} = 15 m/s
t_{up} = 1.5 s
t_{total} = 3.0 s
y at peak = 11.3 m
x = 77.9 m

Calculate:
v_{ix} = 21.2 m/s
v_{iy} = 21.2 m/s
t_{up} = 2.12 s
t_{total} = 4.24 s
y at peak = 22.5 m
x = 89.9 m

Calculate:
v_{ix} = 19.3 m/s
v_{iy} = 23.0 m/s
t_{up} = 2.35 s
t_{total} = 4.69 s
y at peak = 26.9 m
x = 90.4 m

Answer: See table above.
The solutions to all three of these nonhorizontally launched projectile problems involve the use of kinematic equations and an appropriate problemsolving strategy. The kinematic equations and their use in projectile problems are listed and discussed elsewhere. In each of these problems, it is known that a_{x} = 0 m/s/s and a_{y} = 10 m/s/s. The values of v_{ix} and v_{iy} can be determined using trigonometric functions:
v_{ix} = v_{i} * cos(theta)

v_{iy} = v_{i} * sin(theta)

Once v_{ix} and v_{iy} are known, the other unknowns can be calculated. The time up to the peak (t_{up}) can be determined using the equation
v_{fy} = v_{iy} + a_{y }*t
where the v_{fy} = 0 m/s (there is no vertical velocity for a projectile when its at its peak) and a_{y} = 10 m/s/s. Once t_{up} is known, the t_{total} (time to travel the entire trajectory both up and down) can be determined by doubling the t_{up}. The horizontal displacement of the projectile (x) can be computed in the usual way using the equation
x = v_{ix}*t + 0.5*a_{x}*t^{2}
where t is the t_{total} value, a_{x} = 0 m/s/s and v_{ix} was the first value calculated (using the trigonometric functions). Finally, the y at the peak (i.e., the peak height) can be calculated using the equation
y = v_{iy}*t + 0.5*a_{y}*t^{2}
where t is the t_{up} value, a_{y} = 10 m/s/s and v_{iy} was one of the first values calculated (using the trigonometric function). The t value in the equation is t_{up} because the peak height is reached when the projectile has traveled for onehalf of its total time; t_{up} is that time. This method will yield the answers given in the table above.
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62. If a projectile is launched horizontally with a speed of 12.0 m/s from the top of a 24.6meter high building. Determine the horizontal displacement of the projectile.
Answer: x = 27.0 m
This horizontallylaunched projectile problem can be (and should be) solved in the same manner as the solution to #60 above. While #60 is broken down for you into nice steps, this problem is not so userfriendly. It is strongly recommended that you begin by listing known values for each of the variables in the kinematic equations. It is helpful to organize the information into two columns  a column of known horizontal information and a column of known vertical information.
Horizontal Motion
x = ???
v_{ix} = 12.0 m/s
a_{x} = 0 m/s/s (true for all projectiles)

Vertical Motion
y = 24.6 m ( means moving down)
v_{iy} = 0.0 m/s (its launched horizontally)
a_{y} = 9.8 m/s/s (true for all projectiles)

Since three pieces of yinformation are now known, a yequation can be employed to find the time.
y = v_{iy}*t + 0.5*a_{y}*t^{2}
Plugging and chugging the above values into this equation yields a time of 2.25 seconds. Now the t value can be combined with the v_{ix} and a_{x} value and used in an xequation
x = v_{ix}*t + 0.5*a_{x}*t^{2}
to yield the answer 27.0 m. More examples and discussion of these types of projectile problems are discussed elsewhere.
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63. A projectile is launched with an initial speed of 21.8 m/s at an angle of 35.0degrees above the horizontal.
(a) Determine the time of flight of the projectile.
(b) Determine the peak height of the projectile.
(c) Determine the horizontal displacement of the projectile.
Answer: (a) 2.55 s; (b) 7.98 m; (c) 45.7 m
This nonhorizontallylaunched projectile problem can be (and should be) solved in the same manner as the solution to #61 above. While #61 is broken down for you into nicelystructured steps, this problem is not so userfriendly. It is strongly recommended that you begin by resolving the initial velocity and angle into initial velocity components using the equations:
v_{ix} = v_{i} * cos(theta)

v_{iy} = v_{i} * sin(theta)

This yields values of v_{ix} = 17.9 m/s and v_{iy} = 12.5 m/s. Once done, list the known values for each of the variables in the kinematic equations. It is helpful to organize the information into two columns  a column of known horizontal information and a column of known vertical information.
Horizontal Motion
x = ???
v_{ix} = 17.9 m/s (from trig. function)
a_{x} = 0 m/s/s (true for all projectiles)

Vertical Motion
y = 0 m (it rises and falls to original height)
v_{iy} = 12.5 m/s (from trig. function)
a_{y} = 9.8 m/s/s (true for all projectiles)

Since three pieces of yinformation are now known, a yequation can be employed to find the time. One useful equation is
y = v_{iy}*t + 0.5*a_{y}*t^{2}
in which case there will be two solutions: t = 0 s and t = 2.55 s. These two solutions to the equation indicate that the time is 0 s when the vertical displacement (y) is 0 m. This is true before being launched (t = 0 s) and the instant it lands (t=2.55 s). The latter of the two solutions can be used to determine the horizontal displacement (x). Use the equation:
x = v_{ix}*t + 0.5*a_{x}*t^{2}
where t is 2.55 s, a_{x} = 0 m/s/s and v_{ix} was the first value calculated (using the trigonometric functions). Plugging a chugging the above values into this equation yields the answer of 45.7 m.
Finding the vertical displacement at the peak (y_{peak}) demands using the original y equation with a time of 1.28 seconds (t_{up}). This time corresponds to the time for onehalf of the trajectory  the time at which the projectile will be at its highest or peak position. Substituting the v_{iy}, a_{y} and t values into the equation
y_{peak} = v_{iy}*t_{up} + 0.5*a_{y}*t_{up}^{2}
yields a value of 7.98 m for the peak height.
More examples and discussion of these types of projectile problems are discussed elsewhere.
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64. A projectile is launched horizontally from the top of a 45.2meter high cliff and lands a distance of 17.6 meters from the base of the cliff. Determine the magnitude of the launch velocity.
Answer: 5.79 m/s
The best means of starting this problem is to list the known values for each of the variables in the kinematic equations. It is helpful to organize the information into two columns  a column of known horizontal information and a column of known vertical information.
Horizontal Motion
x = 17.6 m (the distance horizontally from cliff base)
v_{ix} = ??? m/s
a_{x} = 0 m/s/s (true for all projectiles)

Vertical Motion
y = 45.2 m (it falls down from the cliff to the ground)
v_{iy} = 0 m/s (it is horizontally launched)
a_{y} = 9.8 m/s/s (true for all projectiles)

Since three pieces of yinformation are now known, a yequation can be employed to find the time. One useful equation is
y = v_{iy}*t + 0.5*a_{y}*t^{2}
in which case there will be two solutions: t = 3.0372 s and t = 3.0372 s. A full parabola which follows the above function would have to locations where the y coordinate is 45.2 m. One would be "forward in time" at 3.0372 seconds; and the other solution is at a location traced "backwards in time" from the launch time. Of course, the positive answer is the one which we need; it can can be used to determine the initial horizontal velocity (v_{ix}). Use the equation:
x = v_{ix}*t + 0.5*a_{x}*t^{2}
where t is 3.0372 s, a_{x} = 0 m/s/s and x = 17.6 m. Plugging and chugging the above values into this equation yields the answer of 5.7948 m/s.
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65. Two physics students stand on the top of their 3.29meter secondstory deck and launch a water balloon from a homemade winger. The balloon is launched upward at a speed of 45.2 m/s and an angle of 39.1 degrees. The balloon lands in a retention pond whose surface is 2.92 meters below grade. Determine the horizontal distance from launch location to landing location.
Answer: 211 m
This is a nonhorizontallylaunched projectile problem in which the initial velocity and launch angle are given. Three initial steps are always wisely taken before starting such a problem. First, determine the initial velocity components (v_{ix} and v_{iy}) using trigonometric functions. Second, construct a diagram of the physical situation. And third, organize known (and unknown) information in an "xy table." These three steps are taken here. Quickly barging into a solution before giving the problem some preanalysis often leads to a wasting of much time and ultimately a lot of confusion.
The initial velocity and angle can be resolved into initial velocity components using the equations:
v_{ix} = v_{i} * cos(theta)

v_{iy} = v_{i} * sin(theta)

This yields values of v_{ix} = 35.077 m/s and v_{iy} = 28.507 m/s.
A diagram of the physical situation is shown.
Now the known values for each of the variables in the kinematic equations is listed in a table using a column for known horizontal information and a column for known vertical information.
Horizontal Motion
x = ???
v_{ix} = 35.077 m/s (from trig. function)
a_{x} = 0 m/s/s (true for all projectiles)

Vertical Motion
y = 5.49 m (from initial + to height to final  height)
v_{iy} = 28.507 m/s (from trig. function)
a_{y} = 9.8 m/s/s (true for all projectiles)

Since three pieces of yinformation are now known, a yequation can be employed to find the time. One useful equation is
y = v_{iy}*t + 0.5*a_{y}*t^{2}
in which case there will be two solutions: t = 0.1867 s and t = 6.004 s. A full parabola which follows the above function would have two locations where the y coordinate is 5.49 m. One location would be "forward in time" at 6.004 seconds; and the other solution is at a location traced "backwards in time" from the launch time. Of course, the positive answer is the one which we need; it can can be used to determine the horizontal displacement (x).
Now use the equation:
x = v_{ix}*t + 0.5*a_{x}*t^{2}
where t = 6.004 s, a_{x} = 0 m/s/s and v_{ix} = 35.077 m/s (as originally calculated using the trigonometric functions). Plugging and chugging the above values into this equation yields the answer of 211 m. (Wow! Those boys had better be careful.)
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66. A place kicker kicks a football from 39.6 meters from the goal posts. The kick leaves the ground with a speed of 24.8 m/s at an angle of 49.6 degrees. The goal posts are 3.10meters high.
(a) Determine the amount by which the kick clears the goal posts.
(b) For this given launch velocity, what is the longest field goal (in yards) which could have been kicked? Assume that the football hits the horizontal crossbar of the posts and bounces through. Given: 1.00 meter = 3.28 feet.
Answer: (a) 13.7 m; (b) 64.7 yds (measured from kick location to goal posts)
(a) In part a of this problem, the task involves finding the height of the ball (y) when it has traveled a distance of 39.6 meters. The height of the goal posts can be subtracted from this value to determine the amount of clearance.
As is the case in all nonhorizontallylaunched projectile problems, it should be begun by resolving the initial velocity and angle into initial velocity components using the equations:
v_{ix} = v_{i} * cos(theta)

v_{iy} = v_{i} * sin(theta)

This yields values of v_{ix} = 16.073 m/s and v_{iy} = 18.886 m/s. Once done, list the known values for each of the variables in the kinematic equations. It is helpful to organize the information into two columns  a column of known horizontal information and a column of known vertical information.
Horizontal Motion
x = 39.6 m (horiz. distance to goal posts)
v_{ix} = 16.073 m/s (from trig. function)
a_{x} = 0 m/s/s (true for all projectiles)

Vertical Motion
y = ??? (we need to calculate this)
v_{iy} = 18.886 m/s (from trig. function)
a_{y} = 9.8 m/s/s (true for all projectiles)

Since three pieces of xinformation are now known, an xequation can be employed to find the time for the football to travel the horizontal distance to the goal posts. One useful equation is
x = v_{ix}*t + 0.5*a_{x}*t^{2}
in which case the time is 2.4637 s. The time can now be combined with a yequations to find the vertical displacement (i.e., height above the ground) when the football has traveled horizontally to the goal posts. Use the equation:
y = v_{iy}*t + 0.5*a_{y}*t^{2}
where t = 2.4637 s, a_{y} = 9.8 m/s/s and v_{iy} = 18.886 m/s. Plugging a chugging the above values into this equation yields the answer of 16.788 m.
When the ball has traveled a horizontal distance of 39.6 m, it is 16.8 m above the ground. The goal posts are 3.10 m high; so the ball clears the goal posts by 13.7 meters.
(b) Part b of this problem can be done in a similar manner. The task will involve first finding the time for the ball to rise to its peak and then fall back down to a height of 3.10 meters. Then the horizontal displacement can be calculated for this time. The final answer will then need to be converted to yards. The same v_{ix} and v_{iy} values can be used. Given the new context of the problem, the value for y is now known and x is unknown. The information can be organized in the usual x and y table.
Horizontal Motion
x = ???
v_{ix} = 16.073 m/s (from trig. function)
a_{x} = 0 m/s/s (true for all projectiles)

Vertical Motion
y = 3.10 m (the height of the goal posts)
v_{iy} = 18.886 m/s (from trig. function)
a_{y} = 9.8 m/s/s (true for all projectiles)

Since three pieces of yinformation are now known, a yequation can be employed to find the time. One useful equation is
y = v_{iy}*t + 0.5*a_{y}*t^{2}
in which case there will be two solutions: t = 0.1718 s and t = 3.6825 s. The first solution corresponds to the first point along the parabola (during the rise of the football) when the football is at a height of 3.10 m and the second solution is the second point along the parabola (during the fall of the football) when the football is at a height of 3.10 m. The second answer can be used to determine the horizontal displacement (x) of the football. Use the equation:
x = v_{ix}*t + 0.5*a_{x}*t^{2}
where t = 3.6825 s, a_{x} = 0 m/s/s and v_{ix} = 16.073 m/s. Plugging a chugging the above values into this equation yields the answer of x = 59.189 m. This x value can be converted to feet by multiplying by the 3.28 ft/m conversion ratio and then converted to yards by dividing by the 3.00 ft/yd conversion ratio. The kicker can kick as 64.7 yard field goal. (In football, it would be referred to as ~47 yard field goal since the goal posts are placed 10 yards behind the goal line and the ball is kicked from about 7 yards behind the line of scrimmage. Field goal distances are are measured from the goal line to the line of scrimmage.)
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67. An airplane starts at Point A and flies 210. km at 311 degrees to Point B. The plane then flies 179 km at 109 degrees to Point C. Finally, the plane flies 228 km at 29 degrees to Point D. Determine the resulting displacement (magnitude and direction) from Points A to D.
Answer: R = 304 km, 24 degrees
Like most vector addition problems, this problem is best begun by the construction of a rough sketch of the physical situation. This sketch is shown below. There are three separate displacements taken by the airplane to result in a single displacement from point A to point D.
The actual solution is best performed using trigonometric functions to determine the x and ycomponents of each displacement vector. These components are then added together to determine the x and ycomponent of the resultant. The work is organized in a table below.

Xcomponent

YComponent

A to B

210. km • cos(311)
= 137.772 km, East

210 km • sin(311)
= 158.489 km, South

B to C

179 km • cos(109)
= 58.277 km, West

179 km • sin(109)
= 169.248 km, North

C to D

228 km • cos(29)
199.413 km, East

228 km • sin(29)
110.537 = km, North

Resultant

278.909 km, East

121.295 km, North

Once the components are known, the Pythagorean theorem can be used to determine the resultant. The resultant has components of 278.909 km, East and 121.295 km, North. The magnitude and direction of the resultant can be determined by adding these components. Since they are at right angles to each other, the magnitude can be determined using the Pythagorean theorem as shown below.
R^{2} = R_{x}^{2} + R_{y}^{2}
R^{2} = (278.909 km)^{2} + (121.295 km)^{2} = 92502.800 km^{2}
R = SQRT(92502.800 km^{2})
R = 304 km
The direction would be stated as the counterclockwise angle of rotation from due East. This is simply the angle Theta. Theta can be determined using the tangent function. The work is shown below.
Tangent(theta) = Ry/Rx
Tangent(theta) = (121.295 km) / (278.909 km) = 0.43489
Theta = Invtan(0.43489)
Theta = 23.5 degrees
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68. Sammy Sosa clubs a homerun which sails 421 feet and lands on an apartment balcony located a vertical distance of 59.0 feet above the level of the ballbat contact location. An observer times the flight to the balcony to take 3.40 seconds.
(a) Determine the velocity (magnitude and angle) at which the ball leaves the bat.
(b) Determine the speed of the ball (in miles/hour) when it lands in the bleachers.
Given: 1.00 m/s = 2.24 mi/hr; 1.00 meter = 3.28 feet.
Answer: (a) vi = 43.7 m/s at 30.2 degrees; (b) 88.3 mi/hr
(a) For any projectile problem, it always a wise idea to begin the solution with a listing of known and unknown information in an "xy table." This is shown below.
Horizontal Motion
x = 421 ft = 128.35 m (horiz. distance to balcony)
v_{ix} = ???
a_{x} = 0 m/s/s (true for all projectiles)
t = 3.40 s (the ball is a projectile for this long)

Vertical Motion
y = 59.0 ft = 17.99 m (vert. distance to balcony)
v_{iy} = ???
a_{y} = 9.8 m/s/s (true for all projectiles)
t = 3.40 s (the ball is a projectile for this long)

Note that the time of flight is known. Time is a scalar quantity and has no directional component associated with it; one cannot refer to the horizontal time or the vertical time. It is listed in both tables since it can be used with kinematic equations for both the x and the ydirection.
Since three pieces of xinformation are known, an xequation can be employed to find the initial horizontal velocity. One useful equation is
x = v_{ix}*t + 0.5*a_{x}*t^{2}
The initial horizontal velocity (v_{ix}) is 37.751 m/s.
There are also three pieces of yinformation known. Thus, a yequation can be used to determine the initial vertical velocity (v_{iy}). A good equation is
y = v_{iy}*t + 0.5*a_{y}*t^{2}
Plugging a chugging the above values into this equation yields an initial vertical velocity (v_{iy}) value of 21.951 m/s.
The ball leaves Sammy Sosa's bat moving upward with a speed of 21.951 m/s and moving horizontally with a speed of 37.751 m/s. These two components of the initial velocity can be used to determine the initial velocity and angle of the baseball after contact with the bat. A diagram is shown at the right. The initial velocity of the ball is represented by the hypotenuse of a right triangle with sides equal to the component values. Thus the Pythagorean theorem can be used to determine the initial velocity of the baseball.
v_{i }^{2} = (v_{ix})^{2} + (v_{iy}) ^{2}
v_{i}^{2} = (37.751 m/s)^{2} + (21.951 m/s)^{2}
v_{i }^{2} = 1906.97 m^{2}/s^{2}
v_{i} = SQRT (1906.97 m^{2}/s^{2} ) = 43.7 m/s
The angle (theta) of the initial velocity can be determined using a trigonometric function. The tangent function is used here.
Tangent(theta) = opposite / adjacent
Tangent(theta) = (21.951 m/s) / (37.751 m/s) = 0.58145
Theta = Invtan (0.58145) = 30.2 degrees
(b) In part (a) of this problem, the initial horizontal velocity was determined to be 37.751 m/s. For projectiles, this horizontal velocity does not change during the flight of the projectile. Thus, the projectile strikes the balcony moving with a final horizontal velocity (v_{fx}) of 37.751 m/s. If the final vertical velocity (v_{fy}) can be determined, then it can be used with the v_{fx} value to determine the final velocity (v_{f}). Several kinematic equations are useable for finding the final vertical velocity (v_{fy}). The following equation will be used:
v_{fy} = v_{iy} + a_{y}•t
v_{fy} = 21.951 m/s + (9.8 m/s/s)•(3.4 s)
v_{fy} = 21.951 m/s  33.32 m/s
v_{fy} = 11.369 m/s
With the x and ycomponents of the final velocity (v_{f}) known, the Pythagorean theorem can be used to determine the final velocity value. A diagram is shown at the right and the calculations are shown below.
v_{f }^{2} = (v_{fx})^{2} + (v_{fy}) ^{2}
v_{f}^{2} = (37.751 m/s)^{2} + (11.369 m/s)^{2}
v_{f }^{2} = 1554.41 m^{2}/s^{2}
v_{f} = SQRT (1554.41 m^{2}/s^{2}) = 39.43 m/s
This value can be converted to miles/hour using the fact that 1.00 m/s = 2.24 mi/hr. The answer to part b is 88.3 mi/hr.
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69. An unfortunate accident occurred on the toll way. A driver accidentally passed through a faulty barricade on a bridge (quite unfortunately). and landed in a pile of hay (quite fortunately). Measurements at the accident scene reveal that the driver plunged a vertical distance of 8.26 meters. The car carried a horizontal distance of 42.1 meters from the location where it left the bridge. If the driver was in a 65 mi/hr speed zone, then determine the amount by which the driver was exceeding the speed limit at the time of the accident. Assume that the contact with the barricade did not slow the car down. (1.00 m/s = 2.24 mi/hr)
Answer: 72.6 mi/hr
This is an example of a horizontallylaunched projectile problem. Like all projectile problems, the best means of starting the problem is to list the known values for each of the variables in the kinematic equations. It is helpful to organize the information into two columns  a column of known horizontal information and a column of known vertical information.
Horizontal Motion
x = 42.1 m (the horizontal distance which is traveled)
v_{ix} = ??? m/s
a_{x} = 0 m/s/s (true for all projectiles)

Vertical Motion
y = 8.26 m (it falls down from the cliff to the ground)
v_{iy} = 0 m/s (it is horizontally launched)
a_{y} = 9.8 m/s/s (true for all projectiles)

Since three pieces of yinformation are now known, a yequation can be employed to find the time. One useful equation is
y = v_{iy}*t + 0.5*a_{y}*t^{2}
in which case there will be two solutions: t = 1.2983 s and t = 1.2983 s. A full parabola which follows the above function would have two locations where the y coordinate is 8.26 m. One would be "forward in time" at 1.2983 seconds; and the other solution is at a location traced "backwards in time" from the launch time. Of course, the positive answer is the one which we need; it can can be used to determine the initial horizontal velocity (v_{ix}). Use the equation:
x = v_{ix}*t + 0.5*a_{x}*t^{2}
where t = 1.2983 s, a_{x} = 0 m/s/s and x = 42.1 m. Plugging a chugging the above values into this equation yields the answer of 32.426 m/s. This is the speed at which the car leaves the bridge at the start of its projectile motion. Converting this to mi/hr involves multiplying by the (2.24 mi/hr) / (1 m/s) conversion ratio. The result is 72.6 mi/hr.
(In reality, the car was traveling faster than this speed since the collision with the guard rail likely slowed the car down before it exited the bridge and began its projectile motion.)
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70. Cupid wishes to shoot an arrow through the open window of a tall building. The window is 32.8 meters above the ground and Cupid stands 63.6 meters from the base of the building. If Cupid aims the arrow at an angle of 51.5 degrees above the horizontal, with what minimum speed must he fire the arrow in order for it to enter the window?
Answer: 32.7 m/s
Here is an example of a nonhorizontally launched projectile problem in which the angle is given but the launch speed is not known. Thus, the x and y components of the initial velocity cannot be found. Nonetheless, expressions relating these components to the initial velocity can still be written and used in the problem.
v_{ix} = v_{i} * cos(theta)

v_{iy} = v_{i} * sin(theta)

The usual procedure of listing the known information in a "xy table" is taken:
Horizontal Motion
x = 63.6 m (horizontal distance to building)
v_{ix} = v_{i} * cos(51.5 deg) = 0.6225 • v_{i}
a_{x} = 0 m/s/s (true for all projectiles)

Vertical Motion
y = 32.8 m (vertical distance from ground to window)
v_{iy} = v_{i} * sin(51.5 deg) = 0.7826 • v_{i}
a_{y} = 9.8 m/s/s (true for all projectiles)

As shown in the table, there are only two pieces of xinformation and two pieces of yinformation given in the problem. Thus, there would seem at first to be insufficient information provided. But as is often the case in a real problem, one can forge ahead using variables in the hopes that there will be a means to introduce another equation which will assist in the solution. So both a horizontal and a vertical displacement equation will be written. (Note that units have been dropped form the solution in order to improve the clarity of the solution.)
Horizontal Displacement
x = v_{ix}*t + 0.5*a_{x}*t^{2}
63.6 = (0.6225 • v_{i} ) • t

Vertical Displacement
y = v_{iy}*t + 0.5*a_{y}*t^{2}
^{}32.8 = (0.7826 • v_{i}) •t + 0.5 • (9.8) • t^{2}

Now we have generated two equations with two unknowns and a solution can be found for the initial velocity of the arrow. Equation 1 is used to generate an expression for t in terms of v_{i}. This expression is then substituted into equation 2 in order to solve for the initial velocity (v_{i}). The work is shown below.
From equation 1: t = (63.6) / (0.6225 • v_{i} )
Substituting into equation 2: 32.8 = (0.7826 • v_{i}) •[(63.6) / (0.6225 • v_{i} )] + 0.5 • (9.8) • [(63.6) / (0.6225 • v_{i} )]^{2}
32.8 = (0.7826 • v_{i}) •[(63.6) / (0.6225 • v_{i} )] + 0.5 • (9.8) • [(63.6) / (0.6225 • v_{i} )]^{2}
32.8 = 79.956  51145.94/(v_{i})^{2}
47.956 = 51145.94/(v_{i})^{2}
(v_{i})^{2} = (51145.94) / (47.956)
(v_{i})^{2} = 1066.52
v_{i} = 32.7 m/s
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71. In a Physics demonstration, a projectile is launched from a height of 1.23 m above the ground with a speed of 10.6 m/s at an angle of 30.0 degrees above the horizontal.
(a) What horizontal distance from the launch location will the projectile land?
(b) With what speed does the projectile land?
Answer: (a) x = 11.7 m; (b) v_{f} = 11.7 m/s
(a) As is the case in all nonhorizontallylaunched projectile problems, it should be begun by resolving the initial velocity (10.6 m/s) and angle (30.0 degrees) into initial velocity components using the equations:
v_{ix} = v_{i} * cos(theta)

v_{iy} = v_{i} * sin(theta)

This yields values of v_{ix} = 9.180 m/s and v_{iy} = 5.30 m/s. Once done, list the known values for each of the variables in the kinematic equations. It is helpful to organize the information into two columns  a column of known horizontal information and a column of known vertical information.
Horizontal Motion
x = ??? (the unknown in part a)
v_{ix} = 9.180 m/s (from trig. function)
a_{x} = 0 m/s/s (true for all projectiles)

Vertical Motion
y = 1.23 m (vert. distance to floor)
v_{iy} = 5.30 m/s (from trig. function)
a_{y} = 9.8 m/s/s (true for all projectiles)

Since three pieces of yinformation are now known, a yequation can be employed to find the time for the projectile to rise and ultimately fall to the floor. One useful equation is
y = v_{iy}*t + 0.5*a_{y}*t^{2}
in which case there are two solutions for the time: t = 1.2780 s and t = 0.1964 s. A full parabola which follows the above function would have two locations where the y coordinate is 1.23 m. One location would be "forward in time" at 1.2780 seconds; and the other solution is at a location traced "backwards in time" from the launch time. Of course, we wish to use the positive time value in our calculations. So t = 1.2780 seconds.
The time can now be combined with a xequations to find the horizontal displacement (x). Use the equation:
x = v_{ix}*t + 0.5*a_{x}*t^{2}
where t = 1.2780 s, a_{x} = 0 m/s/s and v_{ix} = 9.180 m/s. Plugging a chugging the above values into this equation yields the answer of 11.739 meters for the horizontal displacement.
(b) The landing speed (v_{f}) of the projectile can be determined from values of the x and y component of the final velocity. Since the object being analyzed is a projectile, there is no horizontal acceleration and the final horizontal velocity (v_{fx}) is the same as the initial horizontal velocity (v_{ix})  9.180 m/s. The final vertical velocity (v_{fx}) can be determined using the following kinematic equation:
v_{fy} = v_{iy} + a_{y}•t
v_{fy} = 5.3 m/s + (9.8 m/s/s)•(1.2780 s)
v_{fy} = 7.2244 m/s
With the x and ycomponents of the final velocity (v_{f}) known, the Pythagorean theorem can be used to determine the final velocity value. A diagram is shown at the right and the calculations are shown below.
v_{f }^{2} = (v_{fx})^{2} + (v_{fy}) ^{2}
v_{f}^{2} = (9.180 m/s)^{2} + (7.2244 m/s)^{2}
v_{f }^{2} = 136.462 m^{2}/s^{2}
v_{f} = SQRT (136.462 m^{2}/s^{2}) = 11.7 m/s
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72. A car is parked on a cliff overlooking the sea. The cliff is inclined at an angle of 29.0 degrees below the horizontal. The negligent driver leaves the car in neutral and it begins rolling from rest towards the cliff's edge with an acceleration 4.50 m/s/s. The car moves a linear distance of 57.2 m to the edge of the cliff before plunging into the ocean below. The cliff is 42.2 m above the sea.
(a) Find the speed (in m/s) of the car the moment it leaves the cliff.
(b) Find the time (in seconds) it takes the car to drop to the water below the edge of the cliff.
(c) Find the position (in meters) of the car relative to the base of the cliff when it lands in the sea.
Answer: (a) v_{f} = 22.7 m/s; (b) t = 2.02 seconds; (c) x = 40.1 meters
(a) The first task involves using a kinematic equation to determine the speed of the car after accelerating from rest at 4.5 m/s/s for a distance of 57.2 m. The best equation is
v_{f}^{2} = v_{i}^{2} + 2•a•d
v_{f}^{2} = (0 m/s)^{2} + 2•(4.5 m/s^{2})•(57.2 m) = 514.8 m^{2}/s^{2}
v_{f} = SQRT(514.8 m^{2}/s^{2}) = 22.689 m/s
(b) Once the car reaches the edge of the cliff and rolls off, it becomes a projectile with a vertical acceleration of 0 m/s^{2}. The second task involves determining the time of flight of the projectile from the cliff's edge to the water below. Like all nonhorizontally launched projectiles, the starting point is to determine the initial horizontal velocity (v_{ix}) and the initial vertical velocity (v_{iy}). The initial velocity (22.7 m/s) and angle (29.0 degrees) can be resolved into initial velocity components using the equations:
v_{ix} = v_{i} * cos(theta)

v_{iy} = v_{i} * sin(theta)

This yields values of v_{ix} = 19.844 m/s and v_{iy} = 10.999 m/s. Once done, list the known values for each of the variables in the kinematic equations. It is helpful to organize the information into two columns  a column of known horizontal information and a column of known vertical information.
Horizontal Motion
x = ???
v_{ix} = 19.844 m/s (from trig. function)
a_{x} = 0 m/s/s (true for all projectiles)

Vertical Motion
y = 42.2 m (vert. distance to water)
v_{iy} = 10.999 m/s (from trig. function)
a_{y} = 9.8 m/s/s (true for all projectiles)

Since three pieces of yinformation are now known, a yequation can be employed to find the time for the projectile to rise and ultimately fall to the floor. One useful equation is
y = v_{iy}*t + 0.5*a_{y}*t^{2}
in which case there are two solutions for the time: t = 2.0196 s and t = 4.2644 s. A full parabola which follows the above function would have two locations where the y coordinate is 42.2 m. One location would be "forward in time" at 2.0196 seconds; and the other solution is at a location traced "backwards in time" from the launch time. Of course, we wish to use the positive time value in our calculations. So t = 2.0196 seconds.
(c) The time can now be combined with a xequations to find the horizontal displacement (x). Use the equation:
x = v_{ix}*t + 0.5*a_{x}*t^{2}
where t = 2.02 s, a_{x} = 0 m/s/s and v_{ix} = 19.8 m/s. Plugging a chugging the above values into this equation yields the answer of 40.1 meters for the horizontal displacement.
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